\xiti\mylabel{xiti-10}

\begin{xiaotis}

\xiaoti{$k$取什么值时，下列齐次线性方程组有非零解？}
\begin{xiaoxiaotis}

    \twoInLineXxt[16em]{
        $\left\{
            \begin{alignedat}{3}
                kx &+{} &  y & +{}  & z  & = 0, \\
                 x &+{} & 5y & -{}  & 3z & = 0, \\
                2x &+{} &  y &      &    & = 0;
            \end{alignedat}
        \right.$
    }{
        $\begin{cases}
            4x - 5y - 8z = 0, \\
            3x - 4y + kz = 0, \\
            7x - 9y - 5z = 0 \text{。}
        \end{cases}$
    }

\end{xiaoxiaotis}


\xiaoti{下列方程组有没有非零解？如果有，把解集求出来。}
\begin{xiaoxiaotis}

    \twoInLineXxt[16em]{
        $\left\{
            \begin{alignedat}{3}
                2x & +{} & 3y & -{} &  z & = 0, \\
                x  & -{} & 6y & +{} & 2z & = 0, \\
                3x & +{} & 3y & +{} &  z & = 0;
            \end{alignedat}
        \right.$
    }{
        $\left\{
            \begin{alignedat}{3}
                -x &-{} & 2y &+{} &  z & = 0, \\
                2x &+{} & 4y &+{} & 2z & = 0, \\
                x  &+{} & 2y &+{} & 3z & = 0
            \end{alignedat}
        \right.$
    }

\end{xiaoxiaotis}


\xiaoti{已知行列式}
$$\begin{vmatrix*}[r]
    2 & -1 &  3 & 6 \\
    0 &  1 & -6 & 5 \\
    4 &  0 &  3 & 1 \\
    2 &  0 &  1 & 5
\end{vmatrix*} \text{，}$$

\begin{xiaoxiaotis}

    \xiaoxiaoti{把行列式按第三行展开；}

    \xiaoxiaoti{把行列式按第二行展开，并计算出结果。}

\end{xiaoxiaotis}


\xiaoti{利用行列式的性质和展开定理，计算：}
\begin{xiaoxiaotis}

    \renewcommand\arraystretch{1.2}
    \begin{tabular}[t]{*{2}{@{}p{16em}}}
        \xiaoxiaoti{$\begin{vmatrix*}
                1 & 1 &  1 & 1 \\
                1 & 2 &  3 & 4 \\
                1 & 3 &  6 & 10 \\
                1 & 4 & 10 & 20
            \end{vmatrix*}$；}
        & \xiaoxiaoti{$\begin{vmatrix*}
                1 & 2 & 3 & 4 \\
                2 & 3 & 4 & 1 \\
                3 & 4 & 1 & 2 \\
                4 & 1 & 2 & 3
            \end{vmatrix*}$；} \\
    \end{tabular}

    \xiaoxiaoti{$\begin{vmatrix*}
            1 & a_1 & 0 & 0 \\
            -1 & 1 - a_1 & a_2 & 0 \\
            0 & -1 & 1 - a_2 & a_3 \\
            0 & 0 & -1 & 1 - a_3
        \end{vmatrix*}$。}

\end{xiaoxiaotis}


\xiaoti{求证}
$$\begin{vmatrix*}
    \cos\theta & 1 & 0 & 0 \\
    1 & 2\cos\theta & 1 & 0 \\
    0 & 1 & 2\cos\theta & 1 \\
    0 & 0 & 1 & 2\cos\theta
\end{vmatrix*} = \cos4\theta \text{。}
$$


\xiaoti{利用\nameref{klmfz}解下列方程组：}
\begin{xiaoxiaotis}

    \xiaoxiaoti{$\begin{cases}
        x + 2y + 3z - 2w = 6, \\
        2x - y - 2z - 3w = 8, \\
        2y - z + 2w = 1, \\
        2x - 3y + 2z + w = -8;
    \end{cases}$}

    \xiaoxiaoti{$\begin{cases}
        2x + y + 2z - 3w = -3, \\
        x + y - z + 2w = 1, \\
        3x - y + 4z - w = 4, \\
        x + 2y + z - 2w = -5 \text{。}
    \end{cases}$}

\end{xiaoxiaotis}

\end{xiaotis}

